Why is (x^3 - 7x^2 +13x - 6) divisible with (x-2)?
If we look at the general form a polynom P(X) divides Q(x) if and only if Q(x) = P(x) * T(x).Assume P(x) = anxn + an-1xn-1 + ... + a1x + a0. How do the roots of P(x) relate to the roots of Q(x), why?Suppose x1 is a root of P(x), so P(x1) = 0. This means that P(x1)T(x1) = 0.But if x1 is not a root of Q(x) it means that Q(x1) != 0.We conclude that all the roots of P(x) must be roots of Q(x).If (x^3 - 7x^2 +13x - 6) was divisible by (x-2), (x^3 - 7x^2 +13x - 6) = (x-2)L(x). Let's check if all the roots of (x-2) are also roots for (x^3 - 7x^2 +13x - 6). (x-2) has only one root. Now we see that (2^3 - 72^2 +132 - 6) = 0, so all the roots of x-2 are also roots for the given polynom, that's why (x^3 - 7x^2 +13x - 6) is divisible with (x-2).
